Question: The lifespans of turtles in a particular zoo are normally distributed. The average turtle lives $118$ years; the standard deviation is $16.8$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a turtle living less than $151.6$ years.
Explanation: $118$ $101.2$ $134.8$ $84.4$ $151.6$ $67.6$ $168.4$ $95\%$ $2.5\%$ $2.5\%$ We know the lifespans are normally distributed with an average lifespan of $118$ years. We know the standard deviation is $16.8$ years, so one standard deviation below the mean is $101.2$ years and one standard deviation above the mean is $134.8$ years. Two standard deviations below the mean is $84.4$ years and two standard deviations above the mean is $151.6$ years. Three standard deviations below the mean is $67.6$ years and three standard deviations above the mean is $168.4$ years. We are interested in the probability of a turtle living less than $151.6$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the turtles will have lifespans within 2 standard deviations of the average lifespan. The remaining $5\%$ of the turtles will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({2.5\%})$ will live less than $84.4$ years and the other half $({2.5\%})$ will live longer than $151.6$ years. The probability of a particular turtle living less than $151.6$ years is ${95\%} + {2.5\%}$, or $97.5\%$.